Calculate the freezing point of a molar aqueous solution of KCl. (Density of solution = $1.04 g m l^{- 1} K_{f} = 1.86 K k g m o l^{- 1}$ )

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Solution

Molarity of $K C l$ solution $= 1 M$
This means 1 mole (or 74.5 g) of $K C l$ is dissolved in 1 L (or 1000 ml) of solution.
$M a s s o f s o l u t i o n = d e n s i t y o f s o l u t i o n \times v o l u m e o f s o l u t i o n$
$M a s s o f s o l u t i o n = 1.04 \times 1000 = 1040 g$
$M a s s o f s o l v e n t = m a s s o f s o l u t i o n - m a s s o f s o l u t e = 1040 - 74.5 = 965.5 g = 0.9655 k g$
$M o l a l i t y o f s o l u t i o n (m) = \frac{m o l e s o f s o l u t e}{m a s s o f s o l v e n t i n k g}$
$m = \frac{1}{0.9655} = 1.035 m o l k g^{- 1}$
$Δ T_{f} = K_{f} . m$
$(0^{0} C) - (T_{f}) = 1.86 \times 1.035$
$T_{f} = - {1.92}^{0} C = 271.08 K$

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